For \[ A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{bmatrix} \] there is a real orthogonal matrix \(P\) such that \(P^tAP = D\) is diagonal. Find such a diagonal matrix \(D\).
Solution
Because \(A\) is real symmetric, the spectral theorem guarantees a real orthogonal matrix \(P\) for which \(P^tAP\) is diagonal. Its diagonal entries are the eigenvalues of \(A\).
1. Compute the characteristic polynomial.
For \[ A=\begin{pmatrix} 1&2&3\\ 2&3&4\\ 3&4&5 \end{pmatrix}, \] we have \[ \operatorname{tr}(A)=9, \] and the sum of the three principal \(2\times2\) minors is \[ (3-4)+(5-9)+(15-16)=-1-4-1=-6. \] Also \(\det A=0\), since the row differences satisfy \[ (2,3,4)-(1,2,3)=(1,1,1) \] and \[ (3,4,5)-(2,3,4)=(1,1,1). \] Therefore \[ \chi_A(x) =x^3-9x^2-6x =x(x^2-9x-6). \]
2. Read off the diagonal form.
The roots are \[ 0, \qquad \frac{9+\sqrt{105}}2, \qquad \frac{9-\sqrt{105}}2. \] Thus one valid diagonal matrix is \[ \boxed{ D= \begin{pmatrix} \frac{9+\sqrt{105}}2&0&0\\ 0&\frac{9-\sqrt{105}}2&0\\ 0&0&0 \end{pmatrix}.} \] Any permutation of these three diagonal entries is equally valid, with the corresponding permutation of the orthonormal eigenbasis.